Lecture 26

KineticsKineticsElementary Reactions

Reaction OrderMulti-step processes IntermediatesMulti step processes, Intermediates

Steady State Approximation

Reaction OrderReaction Order• The power to which a the concentration of aspecies is raised in the rate law is the reaction

d h ll d h f ll f horder. The overall order is the sum of all of thepowers of all reactants. Examples:Examples:1. v = k[NO]2[O2] First order in O2, Second order in NO, Third order overall.Second order in NO, Third order overall.2. v = k[A]1/2[B]2 Half order in A, Second orderin B, 2 1/2 order overall. • The order does not need to be an integer.

R ti d d l l itReaction order and molecularity

Reaction order is an empirical quantity.Molecularity refers to an elementary reactionas a step in a reaction mechanism.The rate law of an elementary stepcan be deduced directly from the reaction.Unimolecular reactions are first order.A → Products d[A]/dt = -k[A]Bimolecular reactions are second order.A + B → Products d[A]/dt = -k[A][B]

A l 2NO O 2NOAn example: 2NO + O2 2NO2

The formation of nitric oxide is a termolecularreaction, i.e. three molecules must collide inorder for the products to be formed. In thisexample the reaction is also the elementarystep and so the rate law is v = k[NO]2[O2] as stated earlier. In this case the reaction is

f fthird order because of the fact that threemolecules must combine simultaneously.

h hHowever, the reaction stoichiometry may notalways indicate the molecularity of the reaction.

Q ti R ti O dQuestion: Reaction Order

What is the order of the reaction:2 H2 + O2 2 H2O

A. First orderB. Second orderC. Third orderD. Unknown

Q ti R ti O dQuestion: Reaction Order

What is the order of the reaction:2 H2 + O2 2 H2O

A. First orderB. Second orderC. Third orderD. Unknown

Q ti 2 R ti O dQuestion2: Reaction Order

The detailed mechanism of many enzymes hasbeen shown to include a rate determiningstep:

E + S ESWhat is the order of the overall reaction?A. First orderB. Second orderC. Third order

kD. Unknown

Q ti 2 R ti O dQuestion2: Reaction Order

The detailed mechanism of many enzymes hasbeen shown to include a rate determiningstep:

E + S ESWhat is the order of the overall reaction?A. First orderB. Second orderC. Third order

kD. Unknown

P ll l Fi O d R iParallel First Order Reactions

d h• A → B and A → C so that:• d[A]/dt = -(k1+k2)[A]

kA

• d[B]/dt = k1[A]• d[C]/dt = k2[A]

k1B

• Solve for [A] first:k2

[A] = [A]0exp{ – (k 1 + k 2)t }

C thend[B]dt = k 1[A]0exp{ – (k 1 + k 2)t }dt

d[C]dt = k 2[A]0exp{ – (k 1 + k 2)t }

P ll l Fi O d R iParallel First Order Reactions

h l• The solutions are:

kA

[B] = k 1[A]0 1 exp{ (k + k )t }k1B

[B] = 1[ ]0

k 1 + k 21 – exp{ – (k 1 + k 2)t }

[C] = k 2[A]0

k + k1 – exp{ – (k 1 + k 2)t }

• The production of B and C

k2[ ]

k 1 + k 2p{ ( 1 2) }

occurs with a constant proportion:

C[B] = k 1[ ][C] = 1

k 2

Sequential first-order reactionsC i l iConsecutive elementary reactions

• A → B → C rate equations are:k1

A

• d[A]/dt = -k1[A]• d[B]/dt = k1[A] - k2[B]

1B

1 2

• d[C]/dt = k2[B]• Either k1 or k2 can be the rate

k2

1 2limiting step.C

Sequential first-order reactionsC i l iConsecutive elementary reactions

• First solve eqn. for Ak1

A[A] = [A]oe– k 1t

• Substitute into eqn. for B1

B[ ] [ ]o

d[B] = k [A] k [B] = k [A] e– k 1t k [B]k2

[ ]dt = k 1[A] – k 2[B] = k 1[A]oe k 1t – k 2[B]

[B] = k 1[A]o

k 2 – k 1e– k 1t – e– k 2t

• Similarly for CC2 1

[C] = [A]o 1 – 1k 2 – k 1

(k 2e– k 1t – k 1e– k 2t)

l f fPopulations as a function of time

• The population as a function of time is given by the

l ti t th ti lsolutions to the sequential first order reactions.

• The case shown is A - GreenB Y ll

[A](t) - Initial[B](t) I t di t intermediate with k1 = 1.5k2.

• The population of B growsand reaches a maximum and

B - YellowC - Red

[B](t) Intermediate

[C](t) - Final

Time

and reaches a maximum and then decays.

Time

Rate determining stepRate determining step

• If k2 >> k1 then the first (A → B) step becomes the rate-limiting step.g p

• If the A → B step is rate limiting then little orno B will be observed even though it is gformed on the reaction path.

• If k1 >> k2 then the second (B → C) step 1 2 ( ) pbecomes the rate limiting step.

• If the B → C step is rate limiting then there will be a significant build-up of the intermediate state B.

St d t t i tiSteady-state approximation

• Equations representing kinetic networks of more than three states are not soluble analytically.

• One means of pushing the techniques as far as possible using analytical solutions is to set the derivatives of intemediates equal to zero:derivatives of intemediates equal to zero:

d[Intermediate]/dt = 0• The build up of intermediate B shown in the figure• The build-up of intermediate B shown in the figure

(two slides back) implies that the steady state approximation does not work for the system shown pp ythere.

Application of the t d t t i tisteady-state approximation

• The steady state approximation can be applied to the consecutive reaction scheme

k1 k2

A → B → C. if th t ti f B i f i l t tif the concentration of B is fairly constant.

• The result of setting d[B]/dt = 0 is:• The result of setting d[B]/dt = 0 is:k1[A] - k2[B] = 0 and

d[C]/dt = k1[A]. Since d[A]/dt = - k1[A] we see that d[C]/dt d[A]/dt d [C] (1 { k t})d[C]/dt = - d[A]/dt and [C]=(1 - exp{-k1t})

hQuestion: Rate schemes

Which rate equation describes the rate ofdisappearance of A? k2k1

B A C

D

2

k3

1

k-1

A. d[A]/dt = -(k1+ k2 + k3)[A] + k-1[B]B d[A]/dt = (k1+ k2 + k3)[A] - k 1[B]

D

B. d[A]/dt = (k1+ k2 + k3)[A] k-1[B]C. d[A]/dt = - k1[B] - k2[C] - k3[D] + k-1[A] D. d[A]/dt = k1[B] + k2[C] + k3[D] - k-1[A] [ ]/ 1[ ] 2[ ] 3[ ] 1[ ]

hQuestion: Rate schemes

Which rate equation describes the rate ofdisappearance of A? k2k1

B A C

D

2

k3

1

k-1

A. d[A]/dt = -(k1+ k2 + k3)[A] + k-1[B]B d[A]/dt = (k1+ k2 + k3)[A] - k 1[B]

D

B. d[A]/dt = (k1+ k2 + k3)[A] k-1[B]C. d[A]/dt = - k1[B] - k2[C] - k3[D] + k-1[A] D. d[A]/dt = k1[B] + k2[C] + k3[D] - k-1[A] [ ]/ 1[ ] 2[ ] 3[ ] 1[ ]

Example using the steady state i tiapproximation

For the following folding scheme involving asingle intermediate determine the rate ofsingle intermediate determine the rate ofappearance of the folded form F.

k1 k2k1 k2

U ↔ I → F kk-1

d[F]/dt = k2[I]d[I]/dt = k [U] (k + k )[I]d[I]/dt = k1[U] - (k2 + k-1)[I]d[U]/dt = -k1[U] + k-1[I]

Example using the steady state i tiapproximation

Assuming that you have information that allowsyou to use the steady state approximationdetermine the form of d[F]/dt in terms of [U].

k1 k2

U ↔ I → F k-1

A d[F]/dt [U] k k /(k + k )A. d[F]/dt = [U] k1 k2 /(k2 + k-1)B. d[F]/dt = [U] k1 k2

C d[F]/dt = [U] (k + k )/k kC. d[F]/dt = [U] (k2 + k-1)/k1 k2

D. d[F]/dt = [U] (k2 + k-1)

Example using the steady state i tiapproximation

Assuming that you have information that allowsyou to use the steady state approximationdetermine the form of d[F]/dt in terms of [U].

k1 k2

U I FU ↔ I → F k-1

A d[F]/dt = [U] k k /(k + k )A. d[F]/dt = [U] k1 k2 /(k2 + k-1)B. d[F]/dt = [U] k1 k2

C d[F]/dt = [U] (k2 + k )/k k2C. d[F]/dt = [U] (k2 + k-1)/k1 k2

D. d[F]/dt = [U] (k2 + k-1)

Example using the steady state approximation

The equations below show the reasoning behind the solution:

k1 k2

U ↔ I → F k-1

d[I]/dt = k1[U] - (k2 + k-1)[I] = 0k [U] (k k )[I]k1[U] = (k2 + k-1)[I][I] = [U] k1/(k2 + k-1)d[F]/dt k [I] [U] k k /(k + k )d[F]/dt = k2[I] = [U] k1 k2 /(k2 + k-1)

Biexponential kinetics result when the steady-state approximation fails

• We cannot apply the steady state approximation if the concentration of B ppchanges significantly during the process

k1 k2

A → B → C. • In this case we must solve the rate equationsq• In general, for N processes the result will be

kinetics with N exponential time constants. Here: [B](t) = Φ1e-k1t + Φ2e-k2t

Photochemistry

• The intensity, I is measured in power per unit area of photons per second per unit area. The intensity is given in photons/s/m2

i t i / / 2or einsteins/s/m2.• 1 einstein = 1 mole of photons.

T t th i t it i W/ 2 t t• To convert the intensity in W/ m2 to to intensity in number of photon we use the relation: hrelation:

Ephoton = hν = hcλ

Photochemistry

• Beer-Lambert law is:

I = I 10– A A = εlc c = [E]• The concentration is c or [E]• The intensity of absorbed light is I – I

I Io10 , A εlc , c [E]

• The intensity of absorbed light is Io I. • The extinction coefficient ε has units of

M-1cm-1 The exctinction coefficientM 1cm 1. The exctinction coefficient depends on the frequency ε(ω). Often ε is tabulated at the peak of absorption abu a d a p a o abso p ospectrum.

Photochemistry

• Therefore, the rate of excitation of the excited state of molecule E is:

– d[E]dt = φ(Io – I)1000

l = φIo(1 – 10– A)1000l

• The quantum yield is φ.• 1000 converts from cm3 to L1000 converts from cm to L

Photochemistry

• We use the approximation:10– A ≈ 1 – 2.303A

• Thus,10 1 2.303A

d[E] 2303φI ε[E]

• Photoexcitation is a pseudo-first order

– d[E]dt = 2303φIoε[E]

NA

• Photoexcitation is a pseudo-first order process where the excitation rate constant is: 2303φI εs

k = 2303φIoεNA

ll lCompeting or parallel processes

S1• Decay of the singlet S1 statecan occur either radiatively by

hν kf kICfluorescence (kf) or by internal conversion (kIC).• d[S ]/dt (k + k )[S ]

S0

• d[S1]/dt = -(kf + kIC)[S1]• The overall decay rate constantis the sum of the rate constants.0 is the sum of the rate constants. The fluorescence quantum yieldis k

Photoexcitationfollowed by return

Φ =k f

k f + k IC

to the S0 ground state.

ldQuestion: Quantum yield

The observed fluorescence quantum yield of Rhodamine is Φf = 0.7. The observed lifetime is13.0 ns. What is the intrinsic fluorescence lifetime?A. 18.6 nsB 13 0 nsB. 13.0 nsC. 9.1 nsD 0 7 nsD. 0.7 ns

ldQuestion: Quantum yield

The observed fluorescence quantum yield of Rhodamine is Φf = 0.7. The observed lifetime is13.0 ns. What is the intrinsic fluorescence lifetime?A. 18.6 nsB 13 0 nsB. 13.0 nsC. 9.1 nsD 0 7 nsD. 0.7 ns